# chain rule problem

• Nov 12th 2009, 05:43 PM
Evan.Kimia
chain rule problem
Find the derivative of the function. y = 2^3 - x^2

cant get the right answer, i think its partly due to me not knowing the derivative of 2^x.

I did:

outside function = 2^x
outside deriv.= 2^xlog(2) <----this is what i got from an online deriv. calc

inside = 3-x^2
inside deriv = -2x
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• Nov 12th 2009, 05:47 PM
pickslides
Please post the original function in a clear manner.
• Nov 12th 2009, 05:51 PM
Evan.Kimia
• Nov 12th 2009, 05:52 PM
flutterby
I dont undersand the whole inside and outside thing you got going on, but i know that since it is an additive functionjust take the derivative of each term. Like you did in the inside part and put them together.

y'= 3x^2-2x

which folows f(x)= x^(n)

nx^(n-1)
• Nov 12th 2009, 05:52 PM
flutterby
Quote:

Originally Posted by Evan.Kimia

oh sorry i get your question now :)
• Nov 12th 2009, 05:58 PM
flutterby
Quote:

Originally Posted by Evan.Kimia

Take y=2^(3-x^2)

youre probably making a simple mistake

-16*log(2)*x/2^x^2
• Nov 12th 2009, 07:04 PM
pickslides
$y= 2^{3-x^2}$

make $u = 3-x^2 \Rightarrow \frac{du}{dx} = -2x$

and $y= 2^u \Rightarrow \frac{dy}{du} = \ln(2)2^u$

$\frac{dy}{dx} = \frac{dy}{du}\times \frac{du}{dx}$

$\frac{dy}{dx} = \ln(2)2^u\times -2x$

$\frac{dy}{dx} = \ln(2)2^{3-x^2}\times -2x$

$\frac{dy}{dx} = -2\ln(2)x2^{3-x^2}$